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RSS icon Comments on Chris Dodd Sweeps Montlake Caucuses!

1

Correction, Erica ... Dodd 2, not Biden. The Senator from MBNA didn't draw a single vote.

Posted by N in Seattle | January 3, 2008 11:38 AM
2

There's no telling how it would have turned out under Iowa rules, since we didn't proceed with Iowa bargaining.

Additional post hoc analysis here:
http://www.dailykos.com/comments/2008/1/2/231613/1449/167#c167

Posted by RonK, Seattle | January 3, 2008 11:44 AM
3

1: Thanks, fixed it.

2: That's true, but that was the Iowa number announced at the end of the night, different bargaining rules notwithstanding.

Posted by ECB | January 3, 2008 11:47 AM
4

Anybody have any good tips on where to watch the Iowa action unfold tonight?

Posted by Travis | January 3, 2008 11:48 AM
5

--While Goldstein blames “personality issues” for Dodd’s “incredibly poor” showing in the national polls...

Translation: Aside from having zero charisma or supporters, Dodd is the ideal candidate!

Posted by Sigourney Beaver | January 3, 2008 11:54 AM
6

Travis, the Edwards camp is watching at Spitfire in Belltown. Your chance to hobnob with Jeff Smith and Paul Berendt.

The spot for Obama fans was also mentioned last night at Montlake, but I don't recall the place (it's somewhere downtown).

As for me, I'll be watching in an undisclosed location.

Posted by N in Seattle | January 3, 2008 11:55 AM
7

Could someone please explain to me the point of a caucus without a 15% threshold? If I'm a Biden/Dodd/whomever supporter, what's my incentive to bail on my guy and throw my support behind someone else?

Posted by DOUG. | January 3, 2008 12:07 PM
8

Sigh. At least Dodd won something.

Posted by Gitai | January 3, 2008 12:08 PM
9

Doug, your candidate still might not get enough votes to earn a delegate. That's what happened to the backers of Richardson, Kucinich, and Gravel last night. So we swept them all into the Dodd tsunami.

That said, I found the 0% threshold rather disconcerting. But take heart ... the 15% threshold will be back in operation at the county, LD, and CD caucuses.

Posted by N in Seattle | January 3, 2008 12:11 PM
10

@9: Okay, so there is a threshold. And if you have 7 delegates it's 14.3% per delegate, right? How is this much different than the 15% threshold?

Posted by DOUG. | January 3, 2008 12:22 PM
11
Posted by Luigi Giovanni | January 3, 2008 12:29 PM
12
@9: Okay, so there is a threshold. And if you have 7 delegates it's 14.3% per delegate, right? How is this much different than the 15% threshold?

No, that's not right at all. Clinton picked up a delegate with 9.8% of the vote (5 votes out of 51 caucusers). The way the calculations are done in Washington is that each candidate's percentage of the total is multiplied by the number of available delegates (in our case, we chose beforehand to give our "precinct" 7 delegates). Each candidate then gets delegates based on the integer portion of the resulting value. After that, additional delegates are doled out in descending order of the decimal portions.

Last night, the final vote counts and raw delegate counts were Obama 17 / 2.333, Edwards 16 / 2.196, Dodd 13 / 1.784, Clinton 5 / 0.686. Based on integers, that's 2-2-1 for O-E-D. There are two more delegates to be assigned. With the largest decimal portion (.784), Dodd gets one, followed by Clinton with the second largest (.686). At that point, we're out of available delegates, so the final distribution is 2-2-2-1, O-E-D-C.

Posted by N in Seattle | January 3, 2008 12:53 PM
13

Interesting. So if Obama had gotten 22 votes, he'd have 3 delegates (22 / 51 x 7 = 3.02) but if he only gets 21 votes, he'd get just 2 delegates (21 / 51 x 7 = 2.88), right?

So that's the point where you want to convince at least one Kucinich/Biden/whomever to come over to Obama, right?

I guess I'll bring a calculator to my precinct caucus!

Posted by DOUG. | January 3, 2008 1:05 PM
14

Dodd started with 6 votes (1 delegate), but picked up Kucinich votes (2), Gravel (2), and Richardson (1).

Posted by seattlerik | January 3, 2008 1:24 PM
15

Almost got it, Doug. If there were still delegates available after the integer allocation, Obama's .882 decimal portion might be at or near the top of those allocations.

I guess I didn't make it clear that Dodd and Clinton got the "decimal delegates" only because their decimal portions were higher than Edwards's and Obama's. All candidates are rank-ordered for eligibility in the secondary allocation.

Yeah, a calculator will definitely come in handy. I plan to bring my laptop with a spreadsheet all set up to do the calculations.

Posted by N in Seattle | January 3, 2008 1:33 PM
16

Got it. But (in this scenario) if Obama wanted to be GUARANTEED the additional delegate, he'd need the 22nd vote.

Posted by DOUG. | January 3, 2008 2:28 PM
17

...or, I suppose, kill two Biden voters, since 21 / 49 x 7 = 3.

Posted by DOUG. | January 3, 2008 2:32 PM
18

Yes, integers guarantee delegates.

Killing voters wouldn't do it. The total is still 51, based on the initial sign-in. But you deserve credit for thinking outside the (pine) box. :-)

Posted by N in Seattle | January 3, 2008 2:43 PM
19

ECB @ 3 -- Thanks for coming out!

When I announced the Iowa-equivalent results, it was with the announced proviso (which I believe Goldy repeated) that the corresponding Iowa-equivalent bargaining had not occurred.

What anybody heard over the drunken din in the den is another matter.

An Iowa first tally would have yielded 4 delegates for Edwards, 3 for Obama, with (as I also announced) over 30% of attendees left in nonviable factions ... enough to form two entire viable factions, or to thoroughly fuck up the first tally results in a number of other creative and interesting ways.

Posted by RonK, Seattle | January 3, 2008 3:04 PM
20

N @ 18 -- Not exactly. In Iowa, the allocation denomimator is based on the original sign-in count.

In Washington, all caucus fatalities, escapees, and surviving members of nonviable factions are removed from the denominator before allocating delegates. The total count of all viable factions is used as teh allocation denominator.

[Note: W do not endorse anyone killing any caucus participants -- even Biden supporters.]

If our virtual precinct had used a 15% threshold under Washington rules (as applies at later-stage caucus proceedings), the result would have been Dodd 2, Edwards 2, Obama 3.

Posted by RonK, Seattle | January 3, 2008 3:18 PM
21

So does this make Chris Dodd President of Montlake?

Posted by NapoleonXIV | January 3, 2008 4:09 PM
22

I think Dodd will make a great VP.

Posted by Will in Seattle | January 3, 2008 4:21 PM

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